Is there a plane algebraic curve with just 3-fold rotational symmetry, but without reflection symmetry?

I am new to the subject of invariant theory, but the Reynolds operator popped up so I tried to calculated some examples for myself. I computed the invariant polynomials under the cylic group $C_3$ of order $3$, given by $120°$ rotations in the plane.

These polynomials, like

$$\frac 1 4 (x^3-3xy^2), \frac 1 4 (3x^2y-y^3), x^2+y^2, ...$$

and their combinations give rise to nice 120°-symmetric plane curves. However, from what I can judge by looking at the plotted images in Wolframalpha, there always seems to be a reflection symmetry along some axis.

Can we get rid of that? Is there an algebraic curve with just 120°-rotational symmetry but without additional reflection axes?

Of course we can take a union of e.g. appropriately arranged circles expressed in a single equation, so we should just talk about irreducible curves. Is my observation even correct? Maybe there is some obvious counterexample. Thank you

Posting this just to get things started. I haven't checked that this is absolutely irreducible, but it looks promising:

$$x^3 - 3 x y^2 = C + (x^3 - 3 x y^2)^2 + y^3 - 3 y x^2. $$

Built from the invariants that you listed. If we have $C=0$ then there will be a singularity at the origin, and Mathematica has problems drawing it smoothly. Below there is a picture of the real points (I got the impression that you were only interested in real plane curves) with $C=1/24$.

Adding a positive definite term of a high enough degree should give us a compact variant: $$ x^3 - 3 x y^2 + (x^2 + y^2)^4 = \frac1{24} + (x^3 - 3 x y^2)^2 + y^3 - 3 y x^2 $$